Let G be a graph of diameter at least three. Can you find an upper bound on the diameter of the complement of G? Prove your findings!

Let G be a connected graph and sq(G) be a graph which contains all vertices and edges of G and moreover edges joining every pair of vertices that were in G at distance 2. In other words, xy is an edge of sq(G) if and only if the distance of x and y in G is at most 2. Prove that sq(G) is 2-connected.

... to n(n +1)/2 > 1, then w must have at least one edge connecting some node ... of G, so the path v ~ u -> w is in G and connects v and w. Thus G is connected. ...

Hamiltonian Graphs and 2-Connected Graphs. ... Hamiltonian Graphs and 2-Connected Graphs are investigated. The solution is detailed and well presented. ...

... components), there is always a path connecting them, in ... if one is disconnected, then the other is connected. An elementary theorem in graph theory is applied ...

... such that deg_G(v) (greater or equal to) n/2 for every vertex v EV(G). Prove that deleting any vertex of G results in a connected graph (ie G is 2-connected). ...

... The edges that connect the vertices can be lines ... Then, these vertices can be connected using edges to ... the vertices and the weighted edges connecting the cities ...

... Suppose G is connected graph. ... By definition of the connected graph there is a path from any vertex of the graph to another vertex of the graph. ...

... int main(int argc, char **argv) { // This data structure is 5 x 5 by definition Graph g; // Connect the edges g.insertEdge(0,1); // "0" is connected to "1" g ...

... in G such that removal of all the vertices in S , and all the edges that connect to at least one vertex in S , yields either a non-connected graph or a graph...

... that are connected to v in G via an edge. By the definition of graph isomorphism, we know that, for every vertex v1 in V(G), there is an edge connecting v and ...